$g = \gcd(a, b) = \gcd(a, c)$, then $\gcd(b,c) \geq g$, and give an example where $\gcd(b,c) > g$

Question

Let $a, b, c$ be natural numbers and $g = \gcd(a, b) = \gcd(a, c)$, then $\gcd(b,c) \geq g$, and give an example where $\gcd(b,c) > g$.

---

Solution: $g = \gcd(a, b) \implies g | b \implies b=g.m_1$

$g = \gcd(a, c) \implies g | c \implies c=g.m_2$

for some $m_1,m_2 \in \Bbb N$.

Thus $\gcd(b,c) = \gcd(g.m_1,g.m_2) = g. \gcd(m_1,m_2) \geq g$, since $\gcd(m_1,m_2) \geq 1.$

Example: $a = 9, b=6,c=12$. Now $\gcd(a, b) = \gcd(a, c) = 3$, but $\gcd(b,c) = 6 >3.$

Is the solution correct?